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Ted Shifrin
Ted Shifrin
00:04
Three-dimensional? Nah.
robjohn
robjohn
@TedShifrin if it's filled
the surface, as the square of a circle is two dimensional
 
1 hour later…
Ted Shifrin
Ted Shifrin
01:09
@robjohn If it’s filled, your equality fails.
shintuku
shintuku
02:06
@Thorgott thanks for the answers!
@Thorgott little follow up if you have time: supposing I know it is non-empty, don't I have to invoke choice if I don't specify how exactly I'm selecting the element?
Thorgott
Thorgott
you don't, "a non-empty set contains an element" is not the axiom of choice, but a tautology
Ted Shifrin
Ted Shifrin
But picking a particular one? :)
One choice … cheap.
shintuku
shintuku
yeah that's what i'm having trouble with, unless we use the fact every subset has a bijection with some $n \in \mathbb{N}$
(somehow i think we can make a choice function with that)
say we have $X$ infinite, $C$ is the collection of its infinite subsets, we have the constraint that $C$ has a minimal element $C^*$, and we define $S=C^*\backslash \{c\}$, naming $c \in C^*$. Although we know $C^*$ is non-empty, that doesn't imply there is a choice function for $c$, no? unless I'm missing something, and I think we have to use the minimal element characteristic somehow
oh: you're saying a choice function is irrelevant for the act of defining the set
Thorgott
Thorgott
if a set is non-empty, it contains an element (by definition!)
that's all there is to this
choice functions for collections consisting of one set are trivial, but we don't even have to care
shintuku
shintuku
02:22
i'll meditate on this, thanks for the help!
Ted Shifrin
Ted Shifrin
02:34
@Thor: My brain is dying. I just told someone on main that a (smooth) function on $U\subset X$ could be smoothly extended to $X$ for any open $U$. Turns out this was all a red herring, but still.
Thorgott
Thorgott
ha, I did the same thing in here a while ago
Ted Shifrin
Ted Shifrin
My excuse is that I cracked a tooth or something. Excruciating pain.
leslie townes
leslie townes
sorry to hear it ted. best wishes. i have no excuse i'm just dumb.
Ted Shifrin
Ted Shifrin
I’m getting dumber.
And my kitten is growing monstrous.
Dental disasters always happen on weekends.
leslie townes
leslie townes
olivia is a demon right now. she comes into the bedroom in the middle of the night and works her claws into the mattress right next to my wife's head. she also hunts and stalks us around the house.
my last dental disaster was like that. maybe friday 6pm. i couldn't get my regular dentist.
Ted Shifrin
Ted Shifrin
02:49
Sounds like Screech. Plus attacking and biting legs, hands, arms..
leslie townes
leslie townes
i'm covered in bruises from bites. she prefers to bite me and scratch my wife.
Ted Shifrin
Ted Shifrin
LOL
leslie townes
leslie townes
and then for half of the day she'll be very sweet. she sat in my lap when i did several hours of work this afternoon, just purring.
she also never does any of this to our daughter. she'll swat her away with no claws but nothing worse than that, and my daughter is old enough to know what ears back and big pupils mean so that doesn't happen so much.
but olivia will attack us for recreation.
Ted Shifrin
Ted Shifrin
Olivia and Screech will get along famously.
Euler 2
Euler 2
eh
my cat never plays
leslie townes
leslie townes
02:53
some cats are very serene.
olivia will wake us up in the middle of the night tossing her toys around the room.
my wife had never previously owned a cat and was expecting a somewhat less violent experience.
Ted Shifrin
Ted Shifrin
Yup.
Toys or the innocent avocado from the kitchen counter. Nectarines soon to,follow.
My adult cats in GA were beyond this.
leslie townes
leslie townes
livvy will wander into the bedroom, throw her squirrel toy on the bed, clamber across both of our bodies, howl, and then run downstairs.
my first cat was a wonderful cat. he was named balthazar. he had long gray hair and had lost an eye at some point. he just followed us around and hopped up on laps.
Ted Shifrin
Ted Shifrin
Sounds familiar. You forgot attacking body parts to wake us up for sure.
Should be fun tonight with my throbbing tooth.
leslie townes
leslie townes
olivia's favorite move is to paw at my wife's head over and over until she gets sprayed with a spray bottle.
Ted Shifrin
Ted Shifrin
The spray bottles are worthless. Screech licks off the water.
leslie townes
leslie townes
02:59
olivia is also inured to it. it worked for a while.
Ted Shifrin
Ted Shifrin
Told you. They’re soulmates.
leslie townes
leslie townes
tooth pain is the worst pain. i had a lot of periodontal issues in my teens and 20s.
Ted Shifrin
Ted Shifrin
I’ve had more root canals than I can count, and 3 pulled teeth and implants.
leslie townes
leslie townes
i had eight teeth removed because they were rooting as baby teeth. and then there were no replacements, we had no dental insurance, had a lot of bone loss, and then finally bone grafting and implants.
Ted Shifrin
Ted Shifrin
OK, you win.
leslie townes
leslie townes
03:02
my periodontist was a very weird person. she asked if i wanted to see the operation for the bone graft and handed me a mirror. i was high on nitrous so of course i said yes.
it was cool although my face was numb for quite a while because of a nerve that had to be moved to perform the operation.
they tell you the grafts come from a 'tissue bank' or 'bone bank.' they mean cadavers.
Ted Shifrin
Ted Shifrin
Or cat claws.
leslie townes
leslie townes
they somehow chemically denature it so you don't get anything but the bone structure, but oof. and god, it hurts.
Ted Shifrin
Ted Shifrin
They did bone grafts for two of my implants, but that was all ground up …
leslie townes
leslie townes
we're technically zombies.
shintuku
shintuku
03:20
ah! i knew real numbers were weird. the definition using Cauchy sequences of rationals doesn't even require that there be a unique Cauchy sequence defining a specific real number
and we can actually have two non-equal Cauchy sequences of rationals defining a real number
sigh, my time in analysis would have been easier knowing real numbers were themselves limits
robjohn
robjohn
@Koro Take a look here
@leslietownes I had eight baby teeth pulled and 4 permanent teeth pulled (in two operations) to make room for permanent teeth and orthodontia.
Ted Shifrin
Ted Shifrin
03:45
@shintuku what do you think infinite decimals are?
shintuku
shintuku
@TedShifrin the authors all tried to warn me, they were always shoving it my face. but I, human, perhaps too human, could see nothing through the misty fields of pride and avarice
today, i look at my past, and can think nothing else than: i should have listened. they were good to me. i shouldn't have let them down
it is with a heavy heart that i now recognize that the archimedean property is not only a property of the real numbers, and not even the defining property of the set of the real numbers as such, but that, a real number is nothing else than a number to which rationals can accumulate
PM 2Ring
PM 2Ring
@PeterJohn $a_n=2^n+1/n$
Koro
Koro
@robjohn Hi professor Rob, how are you? I saw that and I find it satisfactory. Thank you. But why did you delete your answer? In the answer, my confusion was only the usage of $d(1-\sin t)$ (which is, I think, the same as substitution by $u=1-\sin t$)
robjohn
robjohn
@Koro you mean $\mathrm{d}(1-\cos(t))$?
Koro
Koro
yeah
robjohn
robjohn
03:52
That is just preparing for integration by parts. it chooses the $u$ and $\mathrm{d}v$
I deleted my answer because it did not answer the question of "is my approach correct". Someone pointed it out and since you had seen my answer, I deleted it because it did not answer the question.
The person who complained, did not post an answer, but they did comment.
shintuku
shintuku
but we must rejoice, for from the crumbles of fallen castles rise new kingdoms, and from the ashes rises anew the pheonix
Koro
Koro
@robjohn here, that $d(...)$ has not been used.
but anyways I understood. Thank you so much. I even have a better understanding of Integral comparison test also
robjohn
robjohn
@Koro where is the here to which you are referring?
Koro
Koro
in Integrated Circus, 30 mins ago, by robjohn
$$
\begin{align}
\left|\,\int_x^{x+1}\sin\left(t^2\right)\,\mathrm{d}t\,\right|
&=\left|\,\int_{x^2}^{(x+1)^2}\frac{\sin(t)}{2\sqrt{t}}\,\mathrm{d}t\,\right|\\
&=\left|\,\color{#090}{\left.-\frac{\cos(t)}{2\sqrt{t}}\,\right|_{x^2}^{(x+1)^2}}-\color{#C00}{\int_{x^2}^{(x+1)^2}\frac{\cos(t)}{4t^{3/2}}\,\mathrm{d}t}\,\right|\\
&\le\color{#090}{\frac1{2x}+\frac1{2(x+1)}}+\color{#C00}{\int_{x^2}^{(x+1)^2}\frac1{4t^{3/2}}\,\mathrm{d}t-\int_{x^2}^{(x+1)^2}\frac{1-|\cos(t)|}{4t^{3/2}}\,\mathrm{d}t}\\
&=\frac1{2x}+\frac1{2(x+1)}+\frac1{2x}-\frac1{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{1-|\cos(t)|}{4t^{3/2…
This one
As Ted was saying the other day, I was using Cauchy criteria for proving convergence of the integral, that seems to me now more or less the same as integral comparison test
robjohn
robjohn
there, $\sin(x)\,\mathrm{d}x=-\mathrm{d}\cos(x)$
The integral you quoted there was showing that the integral is strictly less than $\frac1x$
The answer I deleted had a different purpose.
Koro
Koro
04:02
I see.
robjohn
robjohn
The point of your question was to verify a proof using sequential limits verifies a standard limit. My answer was just showing that the standard limit existed.
The answer I posted in Integrated Circus, was referring to a comment you made a couple of days ago.
Ted Shifrin
Ted Shifrin
@shintuku I don't know if this got corrected before. German is the language with plenty of declension (four or five cases). French has none, just subject, direct, and indirect object pronouns.
Koro
Koro
@robjohn yes :)
shintuku
shintuku
yep, in french the function denoted by case in german can be inferred by pronouns and determinants
but you could say some of the pronouns and determinants are declined in the same way the german ones are
for instance, "son chien m'a mordu", m' works like the german "mir": "Sein Hund hat mir gebissen"
PM 2Ring
PM 2Ring
You could say the same about English. Old English (Anglo-Saxon) had a typical Germanic case system, but it faded away, presumably due to the influence of Norman French.
shintuku
shintuku
04:18
and we say "mir" is the dative declination of that pronoun in german
Koro
Koro
@shin: you know German?
shintuku
shintuku
a bit
Koro
Koro
wow great :)
Ted Shifrin
Ted Shifrin
Way more declension in German, just like Latin. Plus three genders.
Koro
Koro
2
Q: Getting strict inequality in $|f(x)|\le \frac 1x$, $x\gt 0$, where $f(x)=\int_x^{x+1} \sin t^2 \, dt$

KoroGiven that $f(x)=\int_x^{x+1} \sin t^2 \, dt$ Substituting $u=t^2$ gives: $f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du$ It follows that $|f(x)|\le\frac 1{2x}+\frac 1{2(x+1)}+\frac 1{2x}-\frac...

real-analysis riemann-integration
Strict inequality can be shown like this also :)
PM 2Ring
PM 2Ring
04:23
Finnish has a whole bunch of cases, but it's not in the Indo-European language family. One of the Python ROs is Finnish, and his prepositions can be a bit ... random. ;)
robjohn
robjohn
@Koro That is pretty much what I have in the other room.
Koro
Koro
yes, I understood that :)
thank you so much :)
robjohn
robjohn
I think mine gives a more convincing demonstration that it needs to be strict, however.
You only need to note that the integrand only vanishes at finitely many points to see that $\int_{x^2}^{(x+1)^2}\frac{1-|\cos(t)|}{4t^{3/2}}\,\mathrm{d}t\gt0$
It is positive on a set of positive measure.
Koro
Koro
we have $\frac {|\cos t|-1}{t^{\frac 32}}\le 0$ and both the functions (on LHS and RHS) are continuous so let $h(x)=\frac {|\cos t|-1}{t^{\frac 32}}- 0$ so if $\int h(x)=0$ then we must have $h(x)=0$ for all x on $[a,b]$ (I could prove it by contradiction and property of integrals $\int_a^b=\int_a^c+\int_c^b$ where $c\in (a,b)$)
your answer has also done the same thing. And I don't know measure (as in measure theory) yet.
robjohn
robjohn
you don't need to know measure theory. a function that is positive except on finitely many points has a positive integral.
I have added the explanation that I was going to post to your question to the answer in my room.
Koro
Koro
04:35
yes, this shows the same thing as you have shown
4 mins ago, by Koro
we have $\frac {|\cos t|-1}{t^{\frac 32}}\le 0$ and both the functions (on LHS and RHS) are continuous so let $h(x)=\frac {|\cos t|-1}{t^{\frac 32}}- 0$ so if $\int h(x)=0$ then we must have $h(x)=0$ for all x on $[a,b]$ (I could prove it by contradiction and property of integrals $\int_a^b=\int_a^c+\int_c^b$ where $c\in (a,b)$)
We want to show that $\int h\lt 0$
and since $h$ is continuous, we can't have $\int h=0$ as that would mean $h=0$ identically on $[x^2,(x+1)^2]$
robjohn
robjohn
yes, you've repeated that at least once.
maybe more
Koro
Koro
:(
But I understand that you conclude the same thing using "vanished on finitely many points" and in the answer the same was done using continuity
robjohn
robjohn
all you need to show is that it does not vanish at one point if you're going to use continuity. However, Martin R's answer does not mention continuity, just that the integrand does not vanish identically, which is not sufficient, unless continuity is cited.
Koro
Koro
He did so in the comments when I asked him how he concluded the final result.
robjohn
robjohn
yeah. That should be brought into the answer.
Comments are not for the meat of the answer.
Edward Evans
Edward Evans
05:18
@shintuku Sein Hund hat mich gebissen
Sorry that wasn't meant to be rude. You correctly mentioned that "mir" is the dative declension of the first person singular pronoun, but a direct object takes (in the vast majority of situations) the accusative case, which here is "mich".
shintuku
shintuku
05:36
oh right my bad
er hat mir ein Geschenk gegeben in that case
Edward Evans
Edward Evans
05:50
Yeah right :)
 
1 hour later…
Evil John Rennie
Evil John Rennie
06:55
Simp >:|
shintuku
shintuku
07:30
nice nice, thinking about real numbers as the completion of the rational line explains the proof structure in real analysis for stuff like supremum proofs
KSLLF
KSLLF
Ought math.stackexchange.com/a/592225/879873 have 14 net upvotes? Will Orrick's comment appears to dispute it.
2. Doesn't math.stackexchange.com/a/2947470 merely duplicate math.stackexchange.com/a/55628 on that same page?
3. Ought math.stackexchange.com/a/1800470/879873 have 4 net upvotes? Will Orrick's comment appears to dispute it.
KSLLF
KSLLF
08:20
4. Can you please answer math.stackexchange.com/questions/5595/taking-seats-on-a-plane/…?
Peter John
Peter John
08:32
@PM2Ring How did you find out?
PM 2Ring
PM 2Ring
09:05
@PeterJohn Well, I generated the terms up to n=10, as rational numbers, and noticed that the denominator is n, which is pretty suspicious. I started working on it algebraically, using similar techniques from working on Fibonacci sequences, and the Pell equation. At first I didn't make much progress, but then I noticed that if $b_n = na_n$ then $\frac{b_{n+1}-1}{b_n-1} = 2 + 2/n$. With a bit of trial & error, I quickly discovered that $(b_n-1)/n = 2^n$
shintuku
shintuku
prove the non-negative rational numbers are dense.
Suppose $a,b,c,d \in \mathbb{N}$ with $d,b \neq 0$ and suppose $\frac a b < \frac c d$. Then,
$0 < \frac c d - \frac a b$
$0 < \frac{c}{2d} - \frac{a}{2b}$
$\frac{a}{b} < \frac{c}{2d} - \frac{a}{2b} + \frac{a}{b}$
$\frac{a}{b} < \frac{c}{2d} + \frac{a}{2b} < \frac{c}{2d} + \frac{c}{2d}=\frac c d$
Since $c, 2d, a, 2b \in \mathbb{N}$, $\frac{c}{2d}$ and \frac{a}{2b}$ are non-negative rationals, and the sum of two non-negative rational numbers is a non-negative rational number. Therefore, we have a non-negative rational number between two arbi…
does anyone here believe me
robjohn
robjohn
@PeterJohn here is how I solved your question
Let $2^nb_n=a_n$, then
$$
2^{n+1}nb_{n+1}-2^{n+1}nb_n+\frac{n+2}{n+1}=0\tag1
$$
Therefore,
$$
\begin{align}
b_{n+1}-b_n
&=-\frac{n+2}{n(n+1)2^{n+1}}\tag{2a}\\
&=-2^{-n-1}\left(\frac2n-\frac1{n+1}\right)\tag{2b}\\
&=\frac{2^{-n-1}}{n+1}-\frac{2^{-n}}{n}\tag{2c}
\end{align}
$$
Thus, $b_n=\frac{2^{-n}}{n}+C$. Since $b_1=\frac32$, we get $C=1$. Therefore,
$$
b_n=\frac{2^{-n}}{n}+1\tag3
$$
Multiply $(3)$ by $2^n$ to get
$$
a_n=\frac1{n}+2^n\tag4
$$
Euler 2
Euler 2
Wtf how what is youtube doing
I accidentally searched for a fullstop
PM 2Ring
PM 2Ring
Well, sure, the mean of $\frac a b$ & $\frac c d$ must lie between them. But also check out $\frac{a+c}{b+d}$ and en.wikipedia.org/wiki/Farey_sequence
Evil John Rennie
Evil John Rennie
@Euler2 no this is not youtube
Euler 2
Euler 2
09:20
Yt is weird
I have heard about elsagate
Evil John Rennie
Evil John Rennie
From russian bot ip source you both have same ip address
Euler 2
Euler 2
Næ
Hæhæ
PM 2Ring
PM 2Ring
@robjohn Nice.
Euler 2
Euler 2
I am russian bot
robjohn
robjohn
@PM2Ring yeah, it is essentially a telescoping sum
Evil John Rennie
Evil John Rennie
09:24
next time remember to use nordvpn and btw this is sponsored by nordvpn
Euler 2
Euler 2
I changed my name because trollbot wasnt nice
expressvpn
Now, time for our sponsor raid shadow legends
2
Evil John Rennie
Evil John Rennie
what do you think about skillshare and brilliant
Euler 2
Euler 2
Download grammarly.
Evil John Rennie
Evil John Rennie
i forgot about that
Evil John Rennie
Evil John Rennie
09:48
I hope to ban my account Euler and russian bot so that my sins will be forgiven :)
As you can see I use this account to upvote Euler 2 account which today earned me 100 rep. These are all fake upvotes.
user91500
user91500
10:13
@EvilJohnRennie You are the person who dunno why apostol's analytic NT book is saying that "it can be read by high school students".
 
1 hour later…
Jam
Jam
11:28
So i want to find the ring of regular functions on punctured plane
$A^{2} - (0) $
now if a polynomial is zeroon the whole plane except the origin should be zero everywhere
and the ring of regular functions is K[x,y]\ I(X)
of all polynomials vanishin on the punctured plane which are only the zero polynomial hence the ring is the whole K[x,y]
whats wrong with my solution? the answer here are quite different
math.stackexchange.com/questions/19884/… this is one answer
Peter John
Peter John
@robjohn @PM2Ring Thanks! I have one more. $\sum_{k=1}^n\frac{k}{(k+1)!} = 1-\frac{1}{(n+1)!}$. I can prove this by induction. But I want to get $1-\frac{1}{(n+1)!}$ by manipulating $\sum_{k=1}^n\frac{k}{(k+1)!}$. Could you help?
Euler 2
Euler 2
12:01
What the heck is rennie doing
Stop him
@EvilJohnRennie ???
Someone upvoted 9 of my posts
Where are mods
Jam
Jam
12:22
phrasing my question somewhat different . Any polynomial function whose zero set is of positive measure then it must be zero everywhere?
Thorgott
Thorgott
yes, the zero set of a non-zero polynomial has measure 0
you can prove this inductively
shintuku
shintuku
are there any useful measures for rational numbers that would more or less tell how well some arbitrary $f:X \to [0,1]\subset \mathbb{Q}$ is filling up the interval?
Someone_Evil
Someone_Evil
12:38
@EvilJohnRennie I don't know what you're trying to do, but if you need sort of multiple accounts, please use the account merge contact on main.
@Euler2 Please just keep calm, and use any appropriate avenues of contact. No mod on the network is gonna suspend account on request (outside of doing site-science I suppose). Please do remember to stay calm and avoid name-calling though, even when something weird is happening. Makes things nicer for everyone involved
@Euler2 Serial voting is generally handled by the script. If isn't reversed in 24 hours, please raise a flag on one of the affected posts. Though for upvotes the worst case here is you having some unearned rep
Thorgott
Thorgott
no, $\mathbb{Q}$ is not a space you can do meaningful measure theory on
the only measures on $\mathbb{Q}$ are series of Dirac measures
shintuku
shintuku
oh well, guess i'll just take differences then
thanks for the answer
PM 2Ring
PM 2Ring
12:57
@PeterJohn Use $\frac{k}{(k+1)!} = \frac1{k!}-\frac1{(k+1)!}$
user193319
user193319
Does anyone know of examples of virtually abelian groups (besides just abelian groups)?
Thorgott
Thorgott
I replied to that yesterday
user193319
user193319
Oh...hmmm...I didn't get a notification...Let me look back...
Ah, yes I was just playing around with direct products; semidirect products being next makes sense.
Peter John
Peter John
@PM2Ring Oh it was an easy stuff. Thanks!
PM 2Ring
PM 2Ring
No worries. :)
Euler 2
Euler 2
13:13
@Someone_Evil okay
Mohcine
Mohcine
14:08
Please, I have question why in proof by induction for showing $P^{n}$ is a stochastic matrix we use base step for n=2 (product of two stochastic matrix) and not n=1 as principal of induction says.
0
Q: Chapter 2 Exercise 2 Question (a) Page 84 Linda J. S. Allen 2010

MohcineExercise 2 Question (a) Page 84 Textbook: An Introduction to Stochastic Processes with Applications to Biology 2nd Edition Linda J. S. Allen 2010 Exercise Suppose $P$ is an $N\times N$ stochastic Matrix (column sums equal one), $P= \begin{pmatrix} p_{11} & p_{12} & \ldots & p_{1N} \\ p_{21} & p...

stochastic-processes solution-verification markov-chains alternative-proof transition-matrix
shintuku
shintuku
14:23
@Mohcine whenever you use induction you can choose whatever first step you want
the induction then works on all subsequent steps
@Mohcine if you absolutely want your induction to start at a 0 step, and need an induction from $n=7$ onwards on $f(n)$ you can define $p(n) = f(7+n)$ and you get your desired induction starting at a 0 step, with $p(0) = f(7)$
Mohcine
Mohcine
14:39
@shintuku Thank you so much I thought that always I have to start with the first step is 0 when N, 1 when N^*, 2 when N^{0,1}, and so on ...
shintuku
shintuku
@Mohcine you can, if you want. just define a function like I just did
Mohcine
Mohcine
@shintuku Thank you so much if I understand well, for our example we can define our function like that $f(n)=p(2+n)$
shintuku
shintuku
yep, so you get your first step $f(0) = p(2+0) = p(2)$
Mohcine
Mohcine
thank you again
shintuku
shintuku
np
Nikhil Kumar Singh
Nikhil Kumar Singh
15:34
Reading this paper by Victor H Moll, I don't understand how does one prove Lemma 6.1. Or is it a standard thing which I should know? It states that $A(z)=\int_0^\infty e^{-zt}B(xt) \mathrm dt$. Then $[z^n]A=n!\times [z^n]B$ where $[z^n]R$ is the coefficient of $z$ in the power series expansion of function $R(z)$.
Sorry for the mistake $A(z)=\int_0^\infty e^{-t}B(xt)\mathrm dt$
Or are there links (other papers, MSE or on MO) for the proof of this lemma?
robjohn
robjohn
15:49
$\int_0^\infty e^{-t}x^nt^n\,\mathrm{d}t=n!x^n$
Multiply by $b_n$ and sum in $n$
Nikhil Kumar Singh
Nikhil Kumar Singh
Thank you, I'm an idiot. I didn't understand this trivial thing.
geocalc33
geocalc33
How do you write $\Bbb C$ take away $1$ in latex?
$(\Bbb C_{\ne 1}, \star)$
okay got it
Rover
Rover
16:18
Any ideas to calculate $ \sum_{r=0}^\infty \frac{r}{3^r} $ ?
leslie townes
leslie townes
one way is to differentiate a power series.
people have worked it out at math.stackexchange.com/questions/2181269/summation-of-n-3n
Rover
Rover
@leslietownes Thanks :) , try to ping please
leslie townes
leslie townes
i'm pretty bad about that.
robjohn
robjohn
16:56
@geocalc33 or $\mathbb{C}\setminus\{1\}$
or even $\mathbb{C}-\{1\}$
@leslietownes or at least try to Bing when you're on the Road...
satan 29
satan 29
can anyone help me with matlab
its a pretty basic program
robjohn
robjohn
better than MatLamb... counting sheep with computer algebra systems
satan 29
satan 29 
t=[0:0.001:10]
vs=4*sin(pi * t)
for i =1:length(vs)
    if(vs(i)<=0.7)
        v(i)=0;
    else
        v(i)=vs(i)-0.7;
    end
end
i wanted to plot(t,v), but I got an error saying v and t are of different sizes
apparently t, vs are of length 10001, which makes sense, but v is of length 1000001, how is that possible?
I can't see how vs and v can differ in size
And also, a friend of mine copy pasted this code and ran it in matlab 2016, and it worked
im using the 2021 version
anyone?
robjohn
robjohn
17:12
@satan29 you don't declare v or set its size. How does Matlab determine a vector size?
satan 29
satan 29
@robjohn you dont need to explicitly declare v
robjohn
robjohn
yes, but there has to be some way for it to set the size.
if you don't declare it, there must be some default
it could be that some default has changed
satan 29
satan 29 
x=[-2 : 0.00001 : 20];
for i=1: length(x)
    if(x(i)>=-2 && x(i)<0)
        y(i)=sqrt(x(i)^2+1);
    else if(x(i)>=0 && x(i)<10)
            y(i)=3*x(i)+1;
        else
            y(i)=9*sin(5*x(i)-50);
        end
    end
end
plot(x,y)
heres another example
i dont declare y explicitlly
however this program worked flawlessly
y and x had the same length
robjohn
robjohn
@satan29 x has 20 elements, and y has a large default, I bet, so accessing an element of y for each x won't run out.
satan 29
satan 29
@robjohn x does not have 20 elements
user193319
user193319
17:18
If $\Gamma$ is a countable, discrete group, what is the dual of $\ell^{\infty}(\Gamma)$?
robjohn
robjohn
what does [-2 : 0.00001 : 20] do then?
satan 29
satan 29
0.00001 is the spacing
robjohn
robjohn
yes and -2 is the start
satan 29
satan 29
it contains an array of elements starting from -2 t0 20, equally spaced, with spacing 0.00001
robjohn
robjohn
oh...
satan 29
satan 29
17:20
so there are 22/0.00001 + 1 elements
robjohn
robjohn
@satan29 yeah, I misread the docs.
satan 29
satan 29
i made the code work by initialising v...but I still dont see how the original one fails despite the x,y one working
shintuku
shintuku
does category theory have something I could use to show dedekind cut reals and cauchy sequence reals are in some way equivalent?
leslie townes
leslie townes
the usual way to do this is to note that both constructions produce a complete archimedean order field, and there is only one of those. it isn't expressly categorical.
51
Q: Who first characterized the real numbers as the unique complete ordered field?

Joel David HamkinsNearly every mathematician nowadays is familiar with the fact that there is up to isomorphism only one complete ordered field, the real numbers. Theorem. Any two complete ordered fields are isomorphic. Proof. $\newcommand\Q{\mathbb{Q}}\newcommand\R{\mathbb{R}}$Let us observe first that every co...

lo.logic real-analysis ho.history-overview
robjohn
robjohn
17:35
@satan29 it looks as if x has fewer elements than y, so plot(x,y) should work. Does the code with v and vs fail or is it only the plot(t,v) that fails?
shintuku
shintuku
@leslietownes neat, thank you!
satan 29
satan 29
@robjohn how does x have fewer elements than y?
plot(a,b) works only if a and b have the same no.of elements
plot (x,y) worked
plot(vs,v) will fail bcs vs and v have different elements..(I dont know how)
robjohn
robjohn
@satan29 you've accessed as many elements of y as length(x), so I assume it must be so
satan 29
satan 29
plot (t,vs) works
@robjohn so that makes all of them doesnt it?
robjohn
robjohn
@satan29 that seems clear since vs was created from t
satan 29
satan 29
17:38
yes, but then v was also created from vs
robjohn
robjohn
@satan29 no, I assume that the array is initialized to some default size and then used unless it is declared or initialized.
satan 29
satan 29
didnt get this point..
robjohn
robjohn
@satan29 no, elements were copied in a loop.
so v was created to some default size, then you set some elements in a loop
satan 29
satan 29
@robjohn well yes, but the loop has i=1 to length (vs)
and we associate all elements of vs to elemennts of v
robjohn
robjohn
try setting v(1)=1 and then v(1000000000)=1
satan 29
satan 29
17:41
this still doesnt explain why the y,x one worked
both are working on the same principle...
one is piecewise with 2 intervals, one is piecewise with 3..
robjohn
robjohn
Oh, I see that Matlab dynamically extends the arrays when an element is set outside the current bounds. So I was wrong about the sizes
satan 29
satan 29
hmm am i going out of bounds somewhere?
robjohn
robjohn
@satan29 unless it is left over from some older value, I don't see how v can be bigger than vs
satan 29
satan 29
@robjohn exactly!!
robjohn
robjohn
was v used earlier and given some huge index values?
satan 29
satan 29
17:56
no
this is the first time the variables t,v and vs were used
robjohn
robjohn
I am just reading the docs, but from what I see, I don't know why v is so big
unless, due to you adding many elements in a loop, Matlab decides to extend the array to a huge size
in its own defense
satan 29
satan 29
I've asked a question on stack overflow. fortunately, they had a matlab tag.
0
Q: Unexpected length of array and plotting error as a result

satan 29I wanted to plot the load voltage across the resistor in series with a diode using matlab. This is a rather simple example involving piecewise functions, however I ran into an unexpected error. t=[0:0.001:10] vs=4*sin(pi * t) for i =1:length(vs) if(vs(i)<=0.7) v(i)=0; else ...

matlab
robjohn
robjohn
The only reason might be that v(i)=0 is being executed a large number of times in a row and Matlab extends the array size to keep from resizing the array over and over. It might be smart enough in the other code to know that each assignment is from an array of a given size and resizes y to match x.
That is the only difference I see. Let's see what the Matlab people say
satan 29
satan 29
should i try v(i)=vs(i)-vs(i) then, Lmao
robjohn
robjohn
18:18
@satan29 try and see...
if it still fails, I am stumped.
I see there is a comment with one of my first thoughts.
unless it is smart enough to use 0 and ignore the size of vs :-(
satan 29
satan 29
@robjohn yeah
Ted Shifrin
Ted Shifrin
@shintuku Prove that there is (up to isomorphism) a unique complete totally ordered field.
leslie townes
leslie townes
we agree again. how's screech?
Ted Shifrin
Ted Shifrin
18:53
Scratching and biting as usual. I have an emergency appt at the dentist this afternoon.
leslie townes
leslie townes
best wishes. i hate going to the dentist.
Thorgott
Thorgott
19:08
alright, let me say something categorical, but let me also stress that it shouldn't be taken too seriously
rain1
rain1
How to evaluate this integral? $\int_0^1 \frac{1}{\sqrt{t(1-t)}} dt$
Thorgott
Thorgott
The Dedekind cut construction in general works to give a functor from the category of posets to its full subcategory of Dedekind complete posets that is left adjoint to the inclusion functor. On the other hand, the Cauchy sequence construction in general works to give a functor from the category of metric spaces to its full subcategory of Cauchy complete metric spaces that is left adjoint to the inclusion functor.
Now one can look at the category of totally ordered, Archmidean fields. Each such field is both a poset and a metric space (the metric being induced by the order) and so we have f…
copper.hat
copper.hat
19:26
don't forget to use some neosporin if you get a dedekind cut.
leslie townes
leslie townes
that's what i'm talking about
Ted Shifrin
Ted Shifrin
@rain1 Complete the square?
rain1
rain1
@TedShifrin, I tried that but I wasn'
t sure what to do next?
Ted Shifrin
Ted Shifrin
@Thor: I think that's what shin deserves.
So what did you get to, @rain1?
copper.hat
copper.hat
Chicago is almost as warm as Ireland.
Ted Shifrin
Ted Shifrin
19:36
You're in Chicago, copper?
rain1
rain1
1/sqrt(1/4 - (t - 1/2)^2)
satan 29
satan 29
@rain1 t= sin^2(x) maybe?
copper.hat
copper.hat
@TedShifrin Just passing through :-)
Ted Shifrin
Ted Shifrin
OK, @rain1 ... that's good. So do you know how to do $1/\sqrt{a^2-u^2}$?
Thorgott
Thorgott
lol
copper.hat
copper.hat
19:37
Covid put paid to the direct SFO flight which is preferrable.
rain1
rain1
i don't
Ted Shifrin
Ted Shifrin
You need a trig substitution. What trig identity does that suggest?
@copper.hat Makes no sense to me.
But I haven't been eating 20 pounds of potatoes a day.
rain1
rain1
I thought we wanted to eliminate trig functions when doing integrals
Ted Shifrin
Ted Shifrin
What makes you think that??!!
copper.hat
copper.hat
@TedShifrin Not enough traffic to justify the flight. At the moment only US citizens & perm. residents can enter.
Ted Shifrin
Ted Shifrin
19:40
Ah, makes sense.
copper.hat
copper.hat
@TedShifrin You would not guess that there is a pandemic based on the sardine packing of the flights.
Ted Shifrin
Ted Shifrin
This is why I have no plans to fly for months and months.
copper.hat
copper.hat
Makes sense.
Ted Shifrin
Ted Shifrin
And one of my doubly-vaccinated good friends in SD (approximately my age) was just hospitalized with the Delta variant and was treated with monoclonal antibodies.
 
1 hour later…
robjohn
robjohn
20:53
@NikhilKumarSingh but do you now see how to get the identity you were asking about?
robjohn
robjohn
21:19
@satan29: did you try the silly v(i)=vs(i)-vs(i) to see if that removed the problem?
since all of the comments/answers say: "initialize the variables" instead of answering the question
schn
schn
21:54
@robjohn , in your extension of the little-o concept -- $\lim\limits_{hu\to0}\frac{f(h,u)}{(hu)^m}=0$ -- you have chosen to replace $x$ under the limit in the definition of $f(x)\in o(g(x))$ by $hu$, but not the $x$ in $f(x)$, i.e. not $f(hu)$. With the definition in mind, it makes it look like $hu$ actually is a variable. On the other hand, if writing $f(hu)$, then all occurrences of $h$ and $u$ are as $hu$.
As you say, it is a one-variable concept. However, since you wrote $f(h,u)$, why not $(h,u)\to (0,0)$?
robjohn
robjohn
I wrote that because the function you gave was not a function of $hu$, but a function where $h$ and $u$ must be separated. The $h$ was constant and the $u$ was being integrated, so it is not a function of $hu$.
this is why you need to understand what little-o is saying and not try to make it into a multi-variable concept before understanding completely what the single variable concept is.
Usually the little-o estimate is used only on a small part of the integral and other bounds need to be used on the other parts of the integral (because the little-o estimate is not usually uniform). I pointed that out in some comment or in the answer, I can't remember where.
user521998
Guys, sorry for saying this here, but I'm starting to learn math to do well on an exam, with the aim of taking an engineering course. That said, sometimes I get confused by some things, and it would be nice to have someone to support me, to guide me through my mistakes. So can someone support me if I need it? I can pay in the future
schn
schn
@robjohn , that clarifies why you wrote $f(h,u)$, but why then $hu\to 0$ in $\lim\limits_{hu\to0}\frac{f(h,u)}{(hu)^m}=0$?
robjohn
robjohn
22:24
Because that was the error term from the Taylor series
schn
schn
Right, I was just about to write.
@robjohn , I'm familiar with uniform convergence, however, could you maybe specify in symbols what you mean when stating that "...$h^3u^4\in o\!\left(u^3\right)$, but not uniformly in $h$ if $h$ gets big..."?
robjohn
robjohn
$\lim\limits_{u\to0}\frac{h^3u^4}{u^3}=0$, however, it requires $u$ much closer to $0$ to make $\frac{h^3u^4}{u^3}$ as small when $h$ is big. That is what non-uniformly means. You can't use the same small $u$ for all $h$.
This is part of the problem with the non-specificity of little-o as a limit equalling $0$
it says the limit is $0$, but not how fast the convergence is
and often the speed of convergence is important
schn
schn
22:43
$\lim\limits_{u\to0}\frac{h^3u^4}{u^3}=h^3\lim\limits_{u\to0}u=h^30$, how can $h$ affect this limit unless it can "equal" positive or negative infinity?
Probably it can't, but only the rate of convergence as you say.
Has Big-O also this non-specificity?
@robjohn , if $h$ is bounded, is the rate of convergence then unaffected?
robjohn
robjohn
23:09
@schn with Big-O, you have a constant, but the constant is not specified either, but you can carry that constant along giving a bit more of a bound than with little-o
@schn the rate of convergence can be better if $h$ is small, but that is usually not a problem.

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